Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int mxs = 100100;
const int inf = -10000000;
int node[mxs << 1];
int prv[mxs << 1];
int cur[mxs];
int k;
void add(int v, int u) {
  node[++k] = u;
  prv[k] = cur[v];
  cur[v] = k;
}
int chain[mxs];
int base[mxs];
int size[mxs];
int heavy[mxs];
int head[mxs];
int P[18][mxs];
int d[mxs];
priority_queue<int> t[mxs << 2];
void dfs(int v, int parent) {
  size[v] = 1;
  P[0][v] = parent;
  for (int i = cur[v]; i; i = prv[i]) {
    int u = node[i];
    if (u == parent) continue;
    d[u] = d[v] + 1;
    dfs(u, v);
    size[v] += size[u];
    if (heavy[v] == -1 || size[heavy[v]] < size[u]) heavy[v] = u;
  }
}
int c;
void dfsHLD(int v, int parent) {
  chain[v] = c;
  if (head[c] == -1) head[c] = v;
  base[v] = k++;
  if (heavy[v] == -1) {
    c++;
    return;
  }
  dfsHLD(heavy[v], v);
  for (int i = cur[v]; i; i = prv[i]) {
    int u = node[i];
    if (u == parent) continue;
    if (u != heavy[v]) dfsHLD(u, v);
  }
}
void update(int v, int l, int r, int pos, int val, bool push) {
  if (l == r) {
    if (push)
      t[v].push(-val);
    else
      t[v].pop();
    return;
  }
  int m = (l + r) >> 1;
  if (pos <= m)
    update(v << 1, l, m, pos, val, push);
  else
    update(v << 1 | 1, m + 1, r, pos, val, push);
  if (t[v].size()) t[v].pop();
  t[v].push(max(t[v << 1].top(), t[v << 1 | 1].top()));
}
int getMin(int v, int l, int r, int tl, int tr) {
  if (tl > tr) return inf;
  if (tl == l && tr == r) return t[v].top();
  int m = (l + r) >> 1;
  return max(getMin(v << 1, l, m, tl, min(m, tr)),
             getMin(v << 1 | 1, m + 1, r, max(m + 1, tl), tr));
}
void buildLCA() {
  for (int i = 1; i < 18; i++)
    for (int j = 1; j < mxs; j++)
      if (P[i - 1][j] != -1) P[i][j] = P[i - 1][P[i - 1][j]];
}
void Heavy_Light_Decompose() {
  memset(heavy, -1, sizeof(heavy));
  memset(P, -1, sizeof(P));
  memset(head, -1, sizeof(head));
  k = 1;
  dfs(1, -1);
  dfsHLD(1, -1);
  buildLCA();
  for (int i = 1; i < mxs << 2; i++)
    for (int j = 0; j < 4; j++) t[i].push(inf);
}
int n;
int getPath(int LCA, int virV) {
  int ans = inf;
  while (true) {
    if (chain[virV] == chain[LCA]) {
      ans = max(ans, getMin(1, 1, n, base[LCA], base[virV]));
      break;
    }
    ans = max(ans, getMin(1, 1, n, base[head[chain[virV]]], base[virV]));
    virV = P[0][head[chain[virV]]];
  }
  return ans;
}
int MeetNode(int u, int v) {
  if (d[u] > d[v]) swap(u, v);
  int virU = u, virV = v;
  for (int i = 17; i >= 0; i--)
    if (P[i][v] != -1 && d[P[i][v]] >= d[u]) v = P[i][v];
  int LCA;
  if (u == v)
    LCA = u;
  else {
    for (int i = 17; i >= 0; i--) {
      if (P[i][v] != -1 && P[i][v] != P[i][u]) {
        v = P[i][v];
        u = P[i][u];
      }
    }
    LCA = P[0][u];
  }
  return max(getPath(LCA, virV), getPath(LCA, virU));
}
int id[mxs];
void Query() {
  vector<int> qt;
  int u, v, a;
  char s[10];
  scanf("%d %d %d", &u, &v, &a);
  for (int i = 0; i < a; i++) {
    int res = MeetNode(u, v);
    if (res == inf) break;
    update(1, 1, n, base[id[-res]], inf, 0);
    qt.push_back(-res);
  }
  printf("%d ", qt.size());
  for (int i = 0; i < qt.size(); i++) printf("%d ", qt[i]);
  puts("");
  for (int i = 0; i < qt.size(); i++) {
    update(1, 1, n, base[id[qt[i]]], qt[i], 1);
  }
}
int m;
void ALL() {
  for (int i = 0, city; i < m; i++) {
    scanf("%d", &city);
    id[i + 1] = city;
    update(1, 1, n, base[city], i + 1, true);
  }
}
int q;
int main() {
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d %d", &v, &u);
    add(v, u);
    add(u, v);
  }
  Heavy_Light_Decompose();
  ALL();
  while (q--) Query();
}